Please help how do i go about solving this?

  1. please help how do i go about solving this??

    undefinedundefinedundefined
    A jogger runs in a straight line with an average
    velocity of 4:7 m/s for 5.1 min, and then with
    an average velocity of 4.1 m/s for 3 min.
    What is her total displacement? Answer in
    units of meters
     
  2. jcsd
  3. use the formula:
    [tex]d = V_{average}t[/tex]
     
  4. and then I subtract the two to find the final displacement
     
  5. dextercioby

    dextercioby 12,324
    Science Advisor
    Homework Helper

    No,you have to add the 2 distances.

    Daniel.
     
  6. Well, it all depends what the direction is. If he runs in the same direction both times, you have to add the 2 displacements to find the total displacement.
     
  7. thanks dude
     
  8. dextercioby

    dextercioby 12,324
    Science Advisor
    Homework Helper

    Even if the runs in the opposite direction on the second time interval,the TOTAL distance would still be gotten by adding the 2.

    Daniel.
     
  9. Aren't the words "distance travelled" and "displacement" different. I though displacement meant distance and direction from the original starting point.
     
  10. for the same question, to get the avg. velocity during the time what should i do? i know it is not just to add the 2 velocities given and divide by 2
     
  11. Average velocity is total distance divided by total time.
     
  12. dextercioby

    dextercioby 12,324
    Science Advisor
    Homework Helper

    I think that u've given yourself the answer. :wink: Besides,since in the problem we're not given the directions (so that we would think it as a vector projection problem),it's natural to assume the simpler version,which is what the teacher would want,right...? :wink: :tongue2: :biggrin:

    Daniel.
     
  13. new question

    A 500-kilogram sports car accelerates uni-
    formly from rest, reaching a speed of 30 me-
    ters per second in 6 seconds.
    During the 6 seconds, the car has traveled
    a distance of how many meters??

    i got 180m is that correct?
    i used : distance = avg. velocity * change in time
    = (30m.s) * 6
    = 180m
     
    Last edited: Feb 6, 2005
  14. ha ha, very funny :wink: :biggrin: :tongue:
     
  15. dextercioby

    dextercioby 12,324
    Science Advisor
    Homework Helper

    Check your numbers again...

    Daniel.
     
  16. Careful... 30m/s is not the average velocity, but rather the final velocity. Remember, average velocity is total displacement divided by the total time, or in this case (if a is constant),
    [tex]\frac{V_i + V_f}{2} = \frac{0+30m/s}{2}[/tex]
     

  17. where did u get the 2 from???
     
  18. Isn't the average the sum of the 2 velocities divided by 2? In this case, the displacement would be: [tex]d = V_{ave}t[/tex]
     
  19. dextercioby

    dextercioby 12,324
    Science Advisor
    Homework Helper

    Kay,guys,let's leave side talks and refer to the initial problem ...

    Daniel.
     
  20. your saying I take 30m/s and divide it by 2?? that is not right either.
     
  21. I's not "side talking". I'm referring to the "new question". Here, average velocity IS involved in finding the distance.

    Anyways, gotta go. I hope I didn't irritate you too much Daniel :biggrin:
     
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